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Question: Use Divergence Theorem to compute $\int \int_D^\ (F.n) dS$ where S is bounded by $z=4-x^2-y^2$, z=0 and z=1, and the force field is $F = (z^3,x^2y,y^2z)$

Attempt to solve: $$div(F) =x^2+y^2 $$ Using cylindrical coordinates $\int \int_D^\ (F.n) dS=\int\int\int_D div(F) dV$ $=\int_0^{2\pi}\int_0^2\int_0^{4-r^2}(z^2r)\,dz\,dr\,d\theta=\frac{32}{3}\pi $ However, the correct answer is 1. What am I doing wrong? Any help is much appreciated.

nova_star
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  • Your integration domain on the plane $;z=0;$ is the annulus between $;x^2+y^2=3;$ and $;x^2+y^2=4;$ . Where does that $;z^2;$ in the integrand come from? I understand the $;r;$ there is the Jacobian for cyl. coordinates...and are you sure the answer is $;1;$ ? – DonAntonio Feb 11 '18 at 21:31
  • I was thinking $$div(F)=x^2+y^2=r^2$$. Yes the study guide says that the answer is 1. Confirmed with the professor as well. – nova_star Feb 11 '18 at 21:37
  • Thanks. I just don't get anything even close to that. Anyway, check your integration domain, as said above. Besides, using cylindrical coordinates, which seems correct in this case, makes almost sure $;\pi;$ will appear in the final result... – DonAntonio Feb 11 '18 at 21:39

1 Answers1

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The correct setting for the integral should be

$$=\int_0^{2\pi}d\theta\int_0^1 dz \int_0^{\sqrt{4-z}}r^3\,dr$$

indeed

  • $z$ varies from 0 to 1
  • for $z$ fixed, $r$ varies from 0 to $\sqrt{4-z}$
  • the divegence is $r^2$
user
  • 154,566
  • This makes sense. I got $$9\pi$$. But not 1. @gimusi , did you get the same answer? It's still not 1 and I am thinking may be the study guide had the wrong answer. – nova_star Feb 11 '18 at 22:00
  • @nova_star I didn't check the result but only the set up, I verify now! – user Feb 11 '18 at 22:01
  • @nova_star here the solution by wolfy https://www.wolframalpha.com/input/?i=int+r%5E3+dr+dz+dt+t%3D0..2pi,+z%3D0..1,r%3D0..(4-z)%5E.5 – user Feb 11 '18 at 22:04
  • I' m pretty sure by the set up ano no one noted mistakes up to now, anyway 1 is pretty strange as result since I think that $\pi$ should appear in the solution. Maybe there is a typo in the guide. – user Feb 11 '18 at 22:09